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borishaifa [10]
3 years ago
8

Find the slope and reduce of p=(3/4,1 1/4) q=(-1/2,-1)

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
7 0
\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%   (a,b)
p&({{ \frac{3}{4}}}\quad ,&{{ \frac{1}{4}}})\quad 
%   (c,d)
q&({{ -\frac{1}{2}}}\quad ,&{{ -1}})
\end{array}

\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{-1-\frac{1}{4}}{-\frac{1}{2}-\frac{3}{4}}\quad \cfrac{\leftarrow LCD=4}{\leftarrow LCD=4}
\\\\\\
\cfrac{\frac{-4-1}{4}}{\frac{-2\cdot 1-3}{4}}\implies \cfrac{\frac{-5}{4}}{\frac{-5}{4}}\implies \cfrac{-5}{4}\cdot \cfrac{4}{-5}\implies 1
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Given: ∆PQR, m∠R = 90° m∠PQR = 75° M ∈ PR , MP = 18 m∠MQR = 60° Find: RQ
tensa zangetsu [6.8K]

Answer:    RQ= 8.99 ( approx)

Step-by-step explanation:

Let MR= x

Since, In triangle, PRQ, tan 75°= \frac{18+x}{RQ}

⇒ RQ=  \frac{18+x}{tan 75^{\circ}}

Now, In triangle MRQ,

tan 60°= \frac{18+x}{RQ}

⇒ RQ=  \frac{x}{tan 60^{\circ}}

On equating both values of RQ,

\frac{18+x}{tan 75^{\circ}}=\frac{x}{tan 60^{\circ}}

⇒\frac{18+x}{x}=\frac{tan 75^{\circ}}{tan 60^{\circ}}

⇒\frac{18+x}{x}=\frac{tan 75^{\circ}}{tan 60^{\circ}}

⇒\frac{18+x}{x}=2.15470053838

⇒18=2.15470053838x-x

⇒x=15.5884572681≈15.60

Thus RQ=8.99999999999≈8.99


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